The table below shows the number of people in an arena t minutes after an event has ended.
T= minutes after event ended: 4 8 12 16 20 24
F(t)= number of people in area: 15221 10649 7375 4859 3468 2285
-Use the regression capability of your graphing calculator to find an exponential equation matching this data. Round values to three decimal places.
F(T)= ___
-use your equation (with values rounded to three decimal places) to help you answer the following questions.
(Note: do not use the data table to answer these questions) Round your answers to the nearest whole number.
- How many people were in the arena 23 minutes after the event ended?
-How many people were in the arena 60 minutes after the event ended?
How many people were in the arena at the exact moment the event ended?

To represent z+9=14, we can start by subtracting 9 from both sides of the equation:

z + 9 - 9 = 14 - 9

Simplifying the left side of the equation gives:

z = 5

Therefore, the solution to the equation z+9=14 is z=5.

Answer:

- 9/16

Step-by-step explanation:

When multiplying fractions, you can multiply them straight across.

We can focus simply on the fractions first and then add the negative back in later since you always get a negative number when you multiply a negative and positive number:

- (3 /4) * (3 / 4)

- (3 * 3) / (4 * 4)

- (9/16)

-9/16

x=15.4 units

First, some definitions before working the problem:

The three standard trigonometric functions, cosine, tangent, and sine, are defined as follows for right triangles:

[tex]sin(\theta)=\dfrac{opposite}{hypotenuse}[/tex]

[tex]cos(\theta)=\dfrac{adjacent}{hypotenuse}[/tex]

[tex]tan(\theta)=\dfrac{opposite}{adjacent}[/tex]

One memorization tactic is "Soh Cah Toa" where the first capital letter represents one of those three trigonometric functions, and the "o" "a" and "h" represent the "opposite" "adjacent" and "hypotenuse" respectively.

The triangle must be a right triangle, or there wouldn't be a "hypotenuse", because the hypotenuse is always across from the right angle.

Working the problem

For the given triangle, the right angle is at the bottom, so the side on top is the hypotenuse.  We know the angle in the upper right corner, so the side across from it with length 4.5, is the opposite side.

For this triangle, the "opposite" leg is known.  Additionally, the "hypotenuse" is unknown and is our "goal to find" side.

Therefore, the two sides of the triangle that are known or are a "goal to find" are the "opposite" & "hypotenuse".

Out of "Soh Cah Toa," the part that uses "o" & "h" is "Soh".  So, the desired function to use for this triangle is the Sine function.

[tex]sin(\theta)=\dfrac{opposite}{hypotenuse}[/tex]

[tex]sin(17^o)=\dfrac{4.5}{x}[/tex]

Multiply both sides by x, and divide both sides by sin(17°)...

[tex]x=\dfrac{4.5}{sin(17^o)}[/tex]

Make sure your calculator is set to degree mode, and calculate:

[tex]x=\dfrac{4.5}{0.2923717047227...}[/tex]

[tex]x=15.391366289249...[/tex] units

Rounded to the nearest tenth...

[tex]x=15.4[/tex] units

The simplified expression of given term  is: -4 + 2 + -2 + -3x = -4 - 3x

Simplification refers to the process of reducing an expression to its simplest form by combining like terms, removing parentheses, and performing any necessary operations such as addition, subtraction, multiplication, and division. The goal of simplification is to make an expression easier to read and work with, and to reveal any patterns or relationships that may not have been obvious in the original expression. Simplification is an important part of solving equations, evaluating expressions, and performing mathematical operations in general.

We can simplify the expression by combining like terms.

Starting with -4, we add 2 to get:

-4 + 2 = -2

Next, we subtract 2 from -2:

-2 - 2 = -4

Finally, we subtract 3x from -4:

-4 - 3x

So the simplified expression is:

-4 + 2 + -2 + -3x = -4 - 3x

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The savings amount balance in the account is $ 467.39

Given data ,

You just opened a savings account and deposited $300 at 3% that you plan on withdrawing in 15 years.

Using the formula FV = PV * (1 + i)^n, where FV is the future value, PV is the present value, i is the interest rate, and n is the number of periods, we can calculate the future value of the savings account after 15 years.

PV = $300 (the initial deposit)

i = 0.03 (the interest rate per period)

n = 15 years (the number of periods)

FV = $300 (1 + 0.03)^15

FV = $300 ( 1.5579 )

FV = $467.39

Hence , after 15 years, the savings account will have a balance of $ 467.39

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Answer: 44%, 26%, less likely

Step-by-step explanation:

just do the math!!

the circumference of the circle with a radius of  [tex]2.5[/tex] meters, approximated using  [tex]\pi = 3.14[/tex], is  [tex]15.7[/tex] meters. Thus, option C is correct.

The circumference of a closed curved object, such a circle or an ellipse, is the length around its perimeter.

The circumference of a circle is the distance around its perimeter. [tex]C = 2r[/tex] , where r is the radius (the distance from the centre of the circle to any point on its border), and pi is a mathematical constant roughly equal to  [tex]3.14159[/tex], is the formula for a circle's circumference.

The formula for the circumference of a circle is [tex]C = 2\pi r[/tex]  , where r is the radius of the circle.

Substituting the given value of the radius, we get:

[tex]C = 2 \times 3.14 \times 2.5 = 15.7[/tex] meters (approx.)

Therefore, the circumference of the circle with a radius of  [tex]2.5[/tex] meters, approximated using π = 3.14, is 15.7 meters.

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The trigonometric ratio cosA = [tex]\frac{3}{5}[/tex] which is in the lowest form.

Trigonometric ratios are the ratios of the sides of a right triangle. The sine, cosine, and tangent are three popular trigonometric ratios. (tan).

The given triangle is a right angle,

To find the cos angle we need to take the ratio of the length of the side which is next to the angle, it is also called an adjacent side to the length of the longest side of the triangle called the hypotenuse.

[tex]cosA = \frac{Length of the side next to the angle}{length of the longest side of the triangle} \\cosA = \frac{Adjacent side}{Hypotenuse} \\cosA = \frac{6}{10}\\ cosA = \frac{3}{5}[/tex]

Therefore [tex]cosA= \frac{3}{5}[/tex]

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Therefore, the correct answer is option C that is LM is reflected over the y-axis to L'M.

In mathematics, a transformation refers to a change in the position, shape, or size of a geometric figure. Transformations can be classified into four types: translation, rotation, reflection, and dilation.

Here,

The transformation of LM to L'M' involves both translation and reflection. To see the translation, we can compare the x- and y-coordinates of L and L', as well as M and M':

The x-coordinate of L' is 5 units more than the x-coordinate of L: -2 = -7 + 5.

The y-coordinate of L' is 2 units less than the y-coordinate of L: -4 = -2 - 2.

The x-coordinate of M' is 5 units more than the x-coordinate of M: 5 = 0 + 5.

The y-coordinate of M' is 2 units less than the y-coordinate of M: 3 = 5 - 2.

Therefore, we can conclude that LM is translated 5 units right and 2 units down to L'M'. This eliminates options OB and OD. To see the reflection, we can compare the x-coordinates of L and M, and their respective x-coordinates in L' and M':

The x-coordinate of L is negative and the x-coordinate of M is positive.

The x-coordinate of L' is negative and the x-coordinate of M' is positive.

Therefore, we can conclude that LM is reflected over the y-axis to L'M'. This eliminates option OA. Therefore, the correct answer is option OC.

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Answer:

Step-by-step explanation:To find the limit of p(x) as x approaches -3, we can first simplify the expression by factoring the numerator:

p(x) = (x^4 - x^3 - 1) / x^2(x + 1)

= [(x - 1)(x^3 + x^2 - x - 1)] / [x^2(x + 1)]

Now, when x approaches -3, the denominator of the fraction becomes zero, which means we have an indeterminate form of the type 0/0. To evaluate the limit, we can use L'Hopital's rule, which states that if we have an indeterminate form of the type 0/0 or infinity/infinity, we can take the derivative of the numerator and denominator separately and then evaluate the limit again.

Taking the derivative of the numerator and denominator, we get:

p'(x) = [(3x^2 - 2x - 1)(x^2 + 2x) - 2(x - 1)(2x + 1)] / [x^3(x + 1)^2]

Now, plugging in x = -3 into the derivative, we get:

p'(-3) = [(3(-3)^2 - 2(-3) - 1)((-3)^2 + 2(-3)) - 2((-3) - 1)(2(-3) + 1)] / [(-3)^3((-3) + 1)^2]

= [28 - 44] / [(-3)^3(-2)^2]

= -16 / 108

= -4 / 27

Since the derivative is defined and nonzero at x = -3, we can conclude that the original limit exists and is equal to the limit of the derivative, which is:

lim x->-3 p(x) = lim x->-3 [(x - 1)(x^3 + x^2 - x - 1)] / [x^2(x + 1)]

= p'(-3)

= -4 / 27

Therefore, the limit of p(x) as x approaches -3 is equal to -4/27.

Answer:

[tex]\lim_{x \to -3}p(x) =-\dfrac{107}{18}[/tex]

Step-by-step explanation:

Given the function [tex]p(x)=\dfrac{x^4-x^3-1}{x^2(x+1)}[/tex]

Let's give the expressions in the numerator and denominator their own function names so they are easy to refer to:

n, for numerator: [tex]n(x)=x^4-x^3-1[/tex]

d, for denominator:  [tex]d(x)=x^2(x+1)[/tex]

So [tex]p(x)=\dfrac{n(x)}{d(x)}[/tex]

Now, we want the limit of p(x) as x goes to -3.

[tex]\lim_{x \to -3}p(x) =\lim_{x \to -3}\dfrac{n(x)}{d(x)}[/tex]

For limits of quotients, it is important to analyze the numerator and the denominator.

Take a moment to observe that inputting -3 into the denominator is defined and does not equal zero:  [tex]d(-3)=(-3)^2((-3)+1)=-18\ne0[/tex]

Also, observe that inputting -3 into the numerator is defined:  [tex]n(-3)=(-3)^4-(-3)^3-1=81+27-1=107[/tex]

Importantly, both functions n & d are polynomials, which are functions that are continuous over [tex]\mathbb{R}[/tex].

Since both functions n & d are continuous, both n & d are defined at [tex]x=-3[/tex], and [tex]d(-3)\ne0[/tex], then the limit of the quotient is the quotient of the limits:

[tex]\lim_{x \to -3}\dfrac{n(x)}{d(x)}=\dfrac{ \lim_{x \to -3}n(x)}{ \lim_{x \to -3}d(x)}[/tex]

From here, again, since n & d are continuous over [tex]\mathbb{R}[/tex] and defined at the limit, [tex]\lim_{x \to -3}n(x)}=n(-3)[/tex] and [tex]\lim_{x \to -3}d(x)}=d(-3)[/tex].

Therefore,

[tex]\lim_{x \to -3}p(x) =\lim_{x \to -3}\dfrac{n(x)}{d(x)}=\dfrac{ \lim_{x \to -3}n(x)}{ \lim_{x \to -3}d(x)}=\dfrac{n(-3)}{d(-3)}=\dfrac{107}{-18}=-\dfrac{107}{18}[/tex]

Answer:

A square's length & width are equal

Step-by-step explanation:

Kamal's shape is not a square, because a square is equilateral (equal length in all sides), but his square is 2:1,

The quadratic function in standard form is f(x) = x² - 4x - 12.

The coefficients of the quadratic function in standard form is calculated as follows;

The quadratic function is in the form of;

f(x) = ax² + bx + c

when x = -4, f(x) = -12

16a - 4b + c = -12

when x = -3, f(x) = -15

9a - 3b + c = -15

when x = -2, f(x) = -16

4a - 2b + c = -16

when x = -1, f(x) = -15

1a - 1b + c = -15

when x = 0, f(x) = -12

0a + 0b + c = -12

c = -12

Simplifying the equations, the value of a and b is calculated as;

16a - 4b + c = -12

9a - 3b + c = -15

4a - 2b + c = -16

a - b + c = -15

16a - 4b = 0

9a - 3b = -3

4a - 2b = -4

a - b = -3

16a = 4b

a = b/4

Substituting this expression for a into the last equation, we get:

b/4 - b = -3

b - 4b = -12

-3b = -12

b = 4

a = b/4

a = 4/4

a = 1

Therefore, the quadratic function in standard form is:

f(x) = x² - 4x - 12

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The value of expression 14a +7+ 5b+ 2a + 10b will be 16a + 15b + 7

Since Expression is defined as the collection of numbers variables and functions by using signs like addition, subtraction, multiplication, and division.

We are given the expression as;

14a +7+ 5b+ 2a + 10b

Combine like terms;

14a + 2a + 10b +7+ 5b

16a + 15b + 7

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A function whose graph is the graph of y = √x, but is shifted to the left 9 units is [tex]y=\sqrt{x+9}[/tex].

In Mathematics and Geometry, the translation of a geometric figure or graph to the left means subtracting a digit from the value on the x-coordinate of the pre-image;

g(x) = f(x + N)

Conversely, the translation a geometric figure upward means adding a digit to the value on the positive y-coordinate (y-axis) of the pre-image; g(x) = f(x) + N.

Based on the information provided, the transformed function should be written as follows:

y = √x

g(x) = (√x + 9)

[tex]y=\sqrt{x+9}[/tex]

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Answer:

1 inch = 0.6 miles

= 9 inches = 0.6 miles*9

= 9 inches = 5.4 miles

Hence, the answer is 5.4 miles.

Please mark me as brainliest...

Answer:

x^1/4 + 8x^1/4 + 12

Step-by-step explanation:

Using the FOIL method, we get x^1/4 + 8x^1/4 + 12

Answer:

The y-intercept of the graph of the equation is -9. This means that the graph crosses the y-axis at the point (0, -9).

Step-by-step explanation:

To solve this question, we need to plug in x = 0 into the given equation and simplify. We get:

y = 6(0 - 1/2)(0 + 3) y = 6(-1/2)(3) y = -9

Therefore, the y-intercept of the graph of the equation is -9. This means that the graph crosses the y-axis at the point (0, -9).

Answer:

it's an easy one

Step-by-step explanation:

1. combine those -->4v+4=16

2. subtract the like terms--> 4v=16-4--> 4v=12

3. divide--> v=12/4-->3

4. Answer--> v=3

Answer:

[tex]c. 1/5[/tex]

Step-by-step explanation:

You have the points (-5, 0) and (0, 1).

To find the slope, b, of the original line, you can use the formula [tex](y_{1}-y_2) /(x_1-x_2)[/tex].

[tex]b = (0-1)/(-5-0) = -1/-5 = 1/5.[/tex]

The slope of a line parallel to the original line would have the same slope as the original line, therefore [tex]1/5.[/tex]

a. To build the airport at the same distance from the center of each town, they should locate the intersection of the perpendicular bisectors of the segments connecting each pair of towns.

b. To build the airport equidistant from each road, locate the intersection of the angle bisectors at each town

The precise location of the airport will be the point where these perpendicular bisectors intersect. This point will be equidistant from the center of each town.

b. To build the airport equidistant from each road, locate the intersection of the angle bisectors at each town. This point is the precise airport location. Equidistant from all connecting roads.

They could find the intersection of the medians of the triangle formed by the three towns. The airport is located at the centroid of a triangle, equidistant from 3 towns at the intersection of medians.

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Answer:

Assuming you meant to write "x^2 - 4x - 12", there are a few equivalent forms that you could use to represent this expression. One common form is:

(x - 6)(x + 2)

Step-by-step explanation:

This is the factored form of the expression, which shows that it can be written as a product of two linear factors. To see why this is true, you can use the distributive property to expand the product:

(x - 6)(x + 2) = x(x + 2) - 6(x + 2) = x^2 + 2x - 6x - 12 = x^2 - 4x - 12

The "Expression" which is considered equivalent to this expression "(3x-1)-2(x+2)' is (c) x-5.

In mathematics, an algebraic expression is a combination of numbers, variables, which are joined by arithmetic operations (such as addition, subtraction, multiplication, and division).

We have to find the equivalent-expression for (3x-1)-2(x+2);

⇒ (3x-1)-2(x+2),

⇒ (3x - 1) - 2x - 4,

Combing the like-terms together in the above expression,

We get,

⇒ 3x - 2x -1 - 4,

⇒ x - 5,

Therefore, the correct equivalent expression is Option (c).

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The given question is incomplete, the complete question is

Which of the following expression is equivalent to (3x-1)-2(x+2)?

(a) x+3

(b) x+1

(c) x-5

(d) x-3

A: The data provides convincing evidence, at α=0.02 level, that the glucose drug is effective in reducing mean glucose level.

B: The 98% confidence interval for the difference in mean reduction of glucose level between placebo and drug groups is 3.5 to 13.7 mg/dL, supporting the effectiveness of the glucose drug

A: To test whether the glucose drug is effective in producing a reduction in mean glucose level, we will use a two-sample t-test with equal variances assuming normality of the differences.

Let μA be the true mean reduction in glucose level for the placebo group and μB be the true mean reduction in glucose level for the glucose drug group. The null hypothesis is H0: μA - μB = 0, and the alternative hypothesis is Ha: μA - μB < 0

Using the given data, the sample mean reduction for the placebo group is 9.7 mg/dL and for the glucose drug group is 18.3 mg/dL. The pooled sample standard deviation is 8.064 mg/dL, and the t-statistic is calculated to be:

t = (xB - xA) / (sP × √(1/nA + 1/nB))

= (18.3 - 9.7) / (8.064 × √(1/10 + 1/10))

= 2.551

where xA and xB are the sample means, sP is the pooled sample standard deviation, and nA and nB are the sample sizes.

B: To create a 98% confidence interval for the difference in the placebo and the new drug, we will use the formula:

CI = (xB - xA) ± tα/2,sP × √(1/nA + 1/nB)

where xA and xB are the sample means, sP is the pooled sample standard deviation, nA and nB are the sample sizes, and tα/2,sP is the t-value corresponding to a 98% confidence level with 18 degrees of freedom.

Using the values from Part A, we have:

CI = (18.3 - 9.7) ± 2.101 × 8.064 × √(1/10 + 1/10)

= 8.6 ± 5.103

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The correct question is:

Twenty middle-aged men with glucose readings between 90 milligrams per deciliter and 120 milligrams per deciliter of blood were selected randomly from the population of similar male patients at a large local hospital. Ten of the 20 men were assigned randomly to group A and received a placebo. The other 10 men were assigned to group B and received a new glucose drug. After two months, posttreatment glucose readings were taken for all 20 men and were compared with pretreatment readings. The reduction in glucose level (Pretreatment reading − Posttreatment reading) for each man in the study is shown here.

Group A (placebo) reduction (in milligrams per deciliter): 12, 8, 17, 7, 20, 2, 5, 9, 12, 6

Group B (glucose drug) reduction (in milligrams per deciliter): 29, 31, 13, 19, 21, 5, 24, 12, 8, 21

Part A: Do the data provide convincing evidence, at the α = 0.02 level, that the glucose drug is effective in producing a reduction in mean glucose level?

Part B: Create and interpret a 98% confidence interval for the difference in the placebo and the new drug.

The net profit margin for the boots is 42.5%.

Net profit margin is the measure of  how much profit is generated as a percentage of revenue. It is the ratio of net profits to revenues.

Revenue from selling the boots is $80.

Total expenses = cost +  operating expenses

The cost is $30, and the operating expenses is 20% of the revenue. That is:

2/100 * $80 = $16.

Total expenses = $30 + $16 = $46

net profit = $80 - $46 = $34

Net profit margin = (Net profit / Revenue) * 100

                            = (34 / 80) * 100

                             = 42.5%

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The calculated value of the median of the A-list 7 members at the gym is 54

From the question, we have the following parameters that can be used in our computation:

10, 64, 52, 46,54,67,54.

When the numbers are sorted in ascending order, we have

Sorted list = 10, 46, 52, 54, 54, 64, 67

The median is the middle number of the sorted list

So, we have

Middle number = 54

Thsis means that

Memdian = 54

Hence. the median of the A-list 7 members at the gym is 54

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The team can sell 50 half dozen boxes and 50 full dozen boxes to raise at least $1000.

We can derive an inequality from the problem statement which is:

8X + 12Y >= 1000

This inequality represents the amount of money the basketball team needs to raise by selling donuts.

X is the number of half dozen boxes sold

Y is the number of full dozen boxes sold.

To find a solution in terms of the given context, we can plug in different values for X and Y that satisfy the inequality. For example:

If the team sells 50 half dozen boxes and 50 full dozen boxes, they will make:

8(50) + 12(50) = $400 + $600 = $1000

So this is a valid solution that meets the fundraising goal.

If the team sells 100 half dozen boxes and 0 full dozen boxes, they will make:

8(100) + 12(0) = $800

This is not enough to meet the fundraising goal, so it is not a valid solution.

If the team sells 0 half dozen boxes and 100 full dozen boxes, they will make:

8(0) + 12(100) = $1200

This is more than the fundraising goal, so it is a valid solution, but the team may not want to sell so many full dozen boxes.

Therefore, one solution in terms of the given context is that the team can sell 50 half dozen boxes and 50 full dozen boxes to raise at least $1000.

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Answer: y = 3/5x - 2/5

Step-by-step explanation: The slope-intercept form is y = mx+b. Hence, solve for y. 3x - 5y = 2.

Move 5y to the right side and move 2 to the left. 3x - 2 = 5y. Divided 5 for all sides: 3/5x - 2/5 = y. Hence, writing in slope-intercept form is y= mx + b, y = 3/5x - 2/5.